Question

Prove the quadratic formula


`ax^2+bx+c=0`

`x = (-b \pm √(b^2-4ac))/(2a)`

Answer 1

Full explanation down here ↓↓↓↓

Explanation:

Hello!

Let's start our derivation:

• 
  `0 = ax^2 + bx + c`
• 
  `-c = ax^2 + bx`

Now, let's complete the square. Set the x² coefficeint to 1 by factoring out a:

• 
  `-c = a(x^2 + (b)/(a)x)`

Complete the square:

• Take the "B" value:
  `\frac ba`
• Divide it by 2:
  `\frac b{2a}`
• Square it:
  `(b^2)/(4a^2)`

Add it to both sides and balance the equation:

• 
  `-c + (ab^2)/(4a^2) = a(x^2 + \frac ba x + (b^2)/(4a^2))` Balance the equation
• 
  `-c + (b^2)/(4a) = a(x + \frac b{2a})^2` Simplify
• 
  `-4ac + b^2 = 4a^2(x+\frac b{2a})^2\\` Multiply by 4a
• 
  `(b^2 - 4ac)/(4a^2) = (x+\frac b{2a})^2` Divide by 4a²
• 
  `\sqrt{ (b^2 - 4ac)/(4a^2) }= \sqrt{(x+\frac b{2a})^2}` Take the square root of both sides
• 
  `(\pm√(b^2 - 4ac))/(2a) = x + \frac b {2a}` Simplify(add plus or minus)
• 
  `(-b \pm √(b^2 - 4ac))/(2a) = x` Simplify

And there you have it! The full proof of the quadratic formula by completing the square.

Answer 2

ax² + bx + c = 0

x = (-b ± √(b² - 4ac))/2a

First, rewrite the first equation so that the first coefficient is 1. Divide everything by a.

(ax² + bx + c = 0)/a =

x² + (b/a)x + (c/a) = 0

Isolate (c/a) by subtracting (c/a) from both sides

x² + (b/a)x + (c/a) (-(c/a) = 0 (- (c/a)

x² + (b/a)x = 0 - (c/a)

Add spaces

x² + (b/a)x = -c/a

Take 1/2 of the middle term's coefficient and square it. Remember that what you add to one side, you add to the other.

x² + (b/a)x + (b/2a)² = -c/a + (b/2a)²

Simplify the left side of the equation.

x² + (b/a)x + (b/2a)² = (x + (b/2a))²

(x + b/2a))² = ((b²/4a²) - (4ac/4a²)) -> ((b² - 4ac)/(4a²))

Take the square root of both sides of the equation

√(x + b/2a))² = √((b²/4a²) - (4ac/4a²))

x + b/(2a) = (±√(b² - 4ac)/2a

Simplify. Isolate the x.

x = -(b/2a) ± (∛b² - 4ac)/2a = (-b ± √(b² - 4ac))/2a

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